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Dodo1

  • one year ago

Find the equation of the tangent line to the curve (a lemniscate) 2(x^2+y^2)2=25(x^2−y^2) at the point (−3,1). The equation of this tangent line can be written in the form y=mx+b where m is:

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  1. Yahoo!
    • one year ago
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    First Differentiate the function to find the slope.

  2. Dodo1
    • one year ago
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    thank you but i dont know process. i am not sure about calculation, particulary differentiation

  3. Dodo1
    • one year ago
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    I mean i know the process

  4. zepdrix
    • one year ago
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    You know the process, but you need assistance with the differentiation portion?

  5. Dodo1
    • one year ago
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    Yes, when it comes to both chain rule and products. I get really confused

  6. zepdrix
    • one year ago
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    \[\large 2(x^2+y^2)^2=25(x^2−y^2)\]So we'll be differentiating this `implicitly`, meaning, `as is`. No solving for y ahead of time. When we execute the chain rule, a few y' will pop out here and there, and we'll want to solve for that variable. \(\large y'(-3,1)=m\). This is the m in our equation \(\large y=mx+b\). Hmm so let's see...

  7. zepdrix
    • one year ago
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    It appears we have no product rule which is nice.

  8. Dodo1
    • one year ago
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    OK, Y7=(-3,1)=m. right?

  9. zepdrix
    • one year ago
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    We should be able to get a numerical value for m, no x's or y's in our m. I'm not really sure what that Y7 is.

  10. Dodo1
    • one year ago
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    Ops my mistake

  11. zepdrix
    • one year ago
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    Taking the derivative of the left side gives us, `taking the derivative of the outermost function first, the square on the brackets`:\[\large 2\cdot2(x^2+y^2)^1\color{royalblue}{\frac{d}{dx}(x^2+y^2)}\]The blue is due to the chain rule. Understand that part?

  12. Dodo1
    • one year ago
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    yes, so far good.

  13. zepdrix
    • one year ago
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    \[\large 2\cdot2(x^2+y^2)\color{royalblue}{(2x+2yy')}\]So we get something like this on the left side.

  14. Dodo1
    • one year ago
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    i lied, I know product rule, power rule etc, but i dont know what the chain rule is

  15. zepdrix
    • one year ago
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    lol :)

  16. zepdrix
    • one year ago
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    mmmm i always have trouble explaining that one. It's very unintuitive.

  17. zepdrix
    • one year ago
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    hmm

  18. Dodo1
    • one year ago
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    could you give me example of chain rule problem maybe?

  19. zepdrix
    • one year ago
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    The definition says this,\[\large \frac{d}{dx}f\left(g(x)\right) \qquad = \qquad f'(g(x))\cdot g'(x)\]When you take the derivative of this, a function within a function, you ~Differentiate the outer function, f in this case. ~Then you make a copy of the inner function, and multiply by it's derivative. Here is a silly example. \[\large \frac{d}{dx}\left(x^2\right)^{4}\] In this example we're thinking of \(\large g(x)=x^2\). And \(\large f(x)=x^4\). If you remember back to doing compositions of functions, you would understand that when we plug g(x) INTO f(x), our f(x), \(\large f(\color{royalblue}{x})=\left(\color{royalblue}{x}\right)^4\) will become, \(\large f(\color{royalblue}{g(x)})=\left(\color{royalblue}{g(x)}\right)^4\) which is \(\large f(\color{royalblue}{x^2})=\left(\color{royalblue}{x^2}\right)^4\). Applying the chain rule gives us,\[\large f'(g(x))=4\left(g(x)\right)^3 g'(x)\]See how the rule tells us to make a copy of the inner function and take it's derivative?\[\large f'(g(x))=4\left(x^2\right)^3 (x^2)'\]\[\large f'(g(x))=4\left(x^2\right)^3 (2x)\]

  20. Dodo1
    • one year ago
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    Oo.. I see, Thank you. It was very clear! really cool :)

  21. zepdrix
    • one year ago
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    If you were to simplify this down, you would get,\[\large f'(g(x))=8x^7\] Which we know is correct, because we could have easily approached this example without using the chain rule.\[\large y=(x^2)^4=x^8\]And we know the derivative will match what we got from the chain rule example.

  22. zepdrix
    • one year ago
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    It's a weird concept :d you're like.. making copies of stuff.. but teachers never use that type of language to describe it. So I dunno lol

  23. Dodo1
    • one year ago
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    no, you explain well. i always get confused and could not tell wether i should use chain or product.

  24. zepdrix
    • one year ago
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    So we applied the derivative of the left side,\[\large \frac{d}{dx}(thing)^2 \qquad \rightarrow \qquad 2(thing)(thing)'\] right? :O

  25. Dodo1
    • one year ago
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    haha thats even more clear haha def, taking note

  26. zepdrix
    • one year ago
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    lol :d

  27. Dodo1
    • one year ago
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    could we back to the first question?

  28. Dodo1
    • one year ago
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    i took note

  29. zepdrix
    • one year ago
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    This one? :o \[\large 2(x^2+y^2)^2=25(x^2−y^2)\]

  30. Dodo1
    • one year ago
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    yes, haha

  31. Dodo1
    • one year ago
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    I see the chain rule now, (x^2+y^2)^2?

  32. zepdrix
    • one year ago
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    \[\large 2(x^2+y^2)\]That's the derivative of the outer function yes? The square on the brackets. But the chain rule is telling us to make a copy of the inside, and take it's derivative.\[\large 2(x^2+y^2)(x^2+y^2)'\]

  33. zepdrix
    • one year ago
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    \[\large 2(x^2+y^2)(2x+2y)\]So we should get somethingggggggg like this, yes?

  34. Dodo1
    • one year ago
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    :O why two in the fron of the bracket? I understand the (2x+2y) tho

  35. zepdrix
    • one year ago
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    You applied the power rule to the outer function \(\large \left(\qquad \right)^2\)

  36. Dodo1
    • one year ago
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    power rule and chain double? :O

  37. zepdrix
    • one year ago
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    Power rule the outside, make a copy of the inside, take derivative of copy.

  38. Dodo1
    • one year ago
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    Ok, i understand

  39. zepdrix
    • one year ago
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    Since we took our derivative with respect to x, any time we differentiate a variable that IS NOT x, like y for example, we have to apply the chain rule, and multiply by y'.

  40. zepdrix
    • one year ago
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    \[\large 2(x^2+y^2)(2x+2y\color{royalblue}{y'})\]See this blue term? We took the derivative of a variable that is not x, namely y, so a y' popped out.

  41. Dodo1
    • one year ago
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    mmmmmm.... Ok. I think :/

  42. Dodo1
    • one year ago
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    so whenever its chain rule i put y' in the end?

  43. zepdrix
    • one year ago
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    No. Let's not think of that portion as chain rule, I think that's just going to confuse you. Whenever we take the derivative of y, we will multiply by y'. We don't throw it on the end, we always attach it to the y that we're taking the derivative of.

  44. Dodo1
    • one year ago
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    right, Ok. i will remember that by my heart.

  45. zepdrix
    • one year ago
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    lol c:

  46. Dodo1
    • one year ago
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    haha

  47. Dodo1
    • one year ago
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    whats nxt stp?

  48. zepdrix
    • one year ago
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    Try taking the derivative of the right side, tell me what you get.

  49. Dodo1
    • one year ago
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    Ok!

  50. zepdrix
    • one year ago
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    Ignore the 25 in front. No chain rule, just take the derivative of the stuff in the brackets.

  51. Dodo1
    • one year ago
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    25(2x-2y)?

  52. zepdrix
    • one year ago
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    close, but you took the derivative of a y, you need to multiply that part by y'

  53. Dodo1
    • one year ago
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    25(2x-2yy')?

  54. Dodo1
    • one year ago
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    its still funny how i have to put y' in the end. is it for implicit derviation?

  55. zepdrix
    • one year ago
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    yes. If we had an equation like this,\[\large y=x^2\]The derivative gives us,\[\large y'=2x\]Ok that makes sense.. The y' is telling us what the derivative of y is. But what if we had something like this?\[\large y^2=x^2\]Taking the derivative,\[\large 2y=2x\] Hmm we don't have a derivative term do we?

  56. Dodo1
    • one year ago
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    Ok, thats cool

  57. Dodo1
    • one year ago
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    so 2(x^2+y~2)(2x+2y)=25(2x-2yy')

  58. zepdrix
    • one year ago
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    Wasn't there a 2 in front of the equation on the left at the start?? Or was that just a typo?

  59. zepdrix
    • one year ago
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    2(x^2+y^2)^2=25(x^2−y^2)

  60. zepdrix
    • one year ago
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    See that 2 at the front? That would give us an extra factor of 2 in our derivative.

  61. Dodo1
    • one year ago
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    the question equation is 2(x^2+y^2)^2=25(x^2-y^2)

  62. Dodo1
    • one year ago
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    so is it 2*2?

  63. zepdrix
    • one year ago
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    yes

  64. Dodo1
    • one year ago
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    \[4(x^2+y^2)(2x+2y)=25(2x-2yy')\]

  65. zepdrix
    • one year ago
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    \[\large 4(x^2+y^2)(2x+2y\color{royalblue}{y'})=25(2x-2yy')\]Woops you missed a y'.

  66. Dodo1
    • one year ago
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    ops! haha thats derivative of the first equation?

  67. zepdrix
    • one year ago
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    yes. We took the derivative of y on each side, so each y will get a y'

  68. Dodo1
    • one year ago
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    Ok,

  69. zepdrix
    • one year ago
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    Hmmm from here, I think it's going to be easier to FIRST plug in your coordinate \(\large (-3,1\;)\) and THEN solve for y'.

  70. Dodo1
    • one year ago
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    plug both number into the equation?

  71. zepdrix
    • one year ago
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    yes, -3 for all the x's. 1 in for all the y's.

  72. zepdrix
    • one year ago
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    and don't plug anything in for y'. That's what we'll be solving for.

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